Problem
You are given an integer array nums
of size n
.
Consider a non-empty subarray from nums
that has the maximum possible bitwise AND.
- In other words, let
k
be the maximum value of the bitwise AND of any subarray ofnums
. Then, only subarrays with a bitwise AND equal tok
should be considered.
Return **the length of the *longest* such subarray**.
The bitwise AND of an array is the bitwise AND of all the numbers in it.
A subarray is a contiguous sequence of elements within an array.
Example 1:
Input: nums = [1,2,3,3,2,2]
Output: 2
Explanation:
The maximum possible bitwise AND of a subarray is 3.
The longest subarray with that value is [3,3], so we return 2.
Example 2:
Input: nums = [1,2,3,4]
Output: 1
Explanation:
The maximum possible bitwise AND of a subarray is 4.
The longest subarray with that value is [4], so we return 1.
Constraints:
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^6
Solution (Java)
class Solution {
public int longestSubarray(int[] nums) {
int max=0;
for(int x:nums)
max=Math.max(max,x);
int maxlen=0;
int count=0;
for(int i=0;i<nums.length;i++)
{
if(nums[i]==max)
{
count++;
}
else{
count=0;
}
maxlen=Math.max(maxlen,count);
}
return maxlen;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).